Triple Integrals in Spherical Coordinates
The spherical coordinates of a point M (x , y , z )ρ , φ , θ
ρ is the length of the radius vector to the point M ;φ is the angle between the projection of the radius vector OM xy -plane and the x -axis;θ is the angle of deviation of the radius vector OM z -axis (Figure 1).
It's important to take into account that the definition of \(\rho\) differs in spherical and cylindrical coordinates.
Figure 1. The spherical coordinates of a point are related to its Cartesian coordinates as follows:
\[x = \rho \cos \varphi \sin \theta ,\;\;y = \rho \sin \varphi \sin \theta ,\;\; z = \rho \cos \theta ,\]
\[\text{where}\;\;\rho \ge 0,\;\; 0 \le \varphi \le 2\pi ,\;\; 0 \le \theta \le \pi .\]
The Jacobian of transformation from Cartesian to cylindrical coordinates is written as
\[
I\left( {\rho ,\varphi ,\theta } \right)
= \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial \rho }}}&{\frac{{\partial x}}{{\partial \varphi }}}&{\frac{{\partial x}}{{\partial \theta }}}\\
{\frac{{\partial y}}{{\partial \rho }}}&{\frac{{\partial y}}{{\partial \varphi }}}&{\frac{{\partial y}}{{\partial \theta }}}\\
{\frac{{\partial z}}{{\partial \rho }}}&{\frac{{\partial z}}{{\partial \varphi }}}&{\frac{{\partial z}}{{\partial \theta }}}
\end{array}} \right|,\]
where the partial derivatives are given by
\[\frac{{\partial x}}{{\partial \rho }} = \cos \varphi \sin \theta,\;\; \frac{{\partial x}}{{\partial \varphi }} = - \rho \sin \varphi \sin \theta,\;\; \frac{{\partial x}}{{\partial \theta }} = \rho \cos \varphi \cos \theta,\;\; \frac{{\partial y}}{{\partial \rho }} = \sin \varphi \sin \theta,\;\; \frac{{\partial y}}{{\partial \varphi }} = - \rho \cos \varphi \sin \theta,\;\; \frac{{\partial y}}{{\partial \theta }} = \rho \sin \varphi \cos \theta,\;\; \frac{{\partial z}}{{\partial \rho }} = \cos \theta,\;\; \frac{{\partial z}}{{\partial \varphi }} = 0,\;\; \frac{{\partial z}}{{\partial \theta }} = -\rho \sin \theta .\]
By expanding the determinant along the second column, we get
\[
I\left( {\rho ,\varphi ,\theta } \right)
= \rho \sin \varphi \sin \theta \cdot \left| {\begin{array}{*{20}{c}}
{\sin \varphi \sin \theta }&{\rho \sin \varphi \cos \theta }\\
{\cos \theta }&{ - \rho \sin \theta }
\end{array}} \right|
+ \rho \cos \varphi \sin \theta \cdot \left| {\begin{array}{*{20}{c}}
{\cos \varphi \sin \theta }&{\rho \cos \varphi \cos \theta }\\
{\cos \theta }&{ - \rho \sin \theta }
\end{array}} \right|
= \rho \sin \varphi \sin \theta \left( { - \rho \sin \varphi \,{{\sin }^2}\theta - \rho \sin \varphi \,{{\cos }^2}\theta } \right)
+ \rho \cos \varphi \sin \theta \left( { - \rho \cos \varphi \,{{\sin }^2}\theta - \rho \cos \varphi \,{{\cos }^2}\theta } \right)
= \rho \sin \varphi \sin \theta \cdot \left( { - \rho \sin \varphi } \right) \cdot 1
+ \rho \cos \varphi \sin \theta \cdot \left( { - \rho \cos \varphi } \right) \cdot 1
= - {\rho ^2}\sin \theta \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)
= - {\rho ^2}\sin \theta .\]
Accordingly, the absolute value of the Jacobian is
\[\left| {I\left( {\rho ,\varphi ,\theta } \right)} \right| = {\rho ^2}\sin \theta .\]
Hence, the formula of change of variables for this transformation is
\[\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} = \iiint\limits_{U'} {f\left( {{\rho \cos \varphi \sin \theta ,\,\rho \sin \varphi \sin \theta,\, \rho \cos \theta }} \right) {\rho ^2}\sin \theta d\rho d\varphi d\theta } .\]
It is easier to calculate triple integrals in spherical coordinates when the region of integration \(U\) is a ball (or some portion of it) and/or when the integrand is a kind of \(f\left( {{x^2} + {y^2} + {z^2}} \right).\)
It is sometimes more convenient to use so-called generalized spherical coordinates , related to the Cartesian coordinates by the formulas
\[x = a\rho \cos \varphi \sin \theta ,\;\; y = b\rho \sin \varphi \sin \theta ,\;\; z = c\rho \cos \theta .\]
In this case the Jacobian is
\[I\left( {\rho ,\varphi ,\theta } \right) = - abc{\rho ^2}\sin \theta .\]
Solved Problems
Example 1.
Evaluate the integral \[\iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz},\] where the region of integration \(U\) is the ball given by the equation \[{x^2} + {y^2} + {z^2} = 25.\]
Solution.
As the region \(U\) is a ball and the integrand is expressed by a function depending on \(f\left( {{x^2} + {y^2} + {z^2}} \right),\) we can convert the triple integral to spherical coordinates. Make the substitution:
\[x = \rho \cos \varphi \sin \theta ,\;\;y = \rho \sin \varphi \sin \theta ,\;\; z = \rho \cos \theta ,\]
The new variables range within the limits:
\[0 \le \rho \le 5,\;\; 0 \le \varphi \le 2\pi ,\;\; 0 \le \theta \le \pi .\]
Taking into account the Jacobian \({\rho ^2}\sin \theta,\) we can write the integral as follows:
\[I = \iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz} = \iiint\limits_{U'} {\rho \cdot {\rho ^2}\sin \theta d\rho d\varphi d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \int\limits_0^\pi {\sin \theta d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right] = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left( { - \cos \pi + \cos 0} \right) = 2\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } = 2\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^4}}}{4}} \right)} \right|_0^5} \right] = 2\int\limits_0^{2\pi } {d\varphi } \cdot \frac{{{5^4}}}{4} = \frac{{625}}{2}\int\limits_0^{2\pi } {d\varphi } = \frac{{625}}{2} \cdot 2\pi = 625\pi .\]
Example 2.
Calculate the integral \[\iiint\limits_U {{e^{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}}dxdydz} ,\] where the region \(U\) is the unit ball \[{x^2} + {y^2} + {z^2} \le 1.\]
Solution.
This ball is centered at the origin. Hence, the region of integration \(U\) in spherical coordinates is described by the inequalities
\[0 \le \rho \le 1,\;\; 0 \le \varphi \le 2\pi ,\;\; 0 \le \theta \le \pi .\]
Writing the integral in spherical coordinates, we have
\[I = \iiint\limits_U {{e^{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}}dxdydz} = \iiint\limits_{U'} {{e^{{{\left( {{\rho ^2}} \right)}^{\frac{3}{2}}}}}{\rho ^2}\sin \theta d\rho d\varphi d\theta } = \iiint\limits_{U'} {{e^{{\rho ^3}}}{\rho ^2}\sin \theta d\rho d\varphi d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{e^{{\rho ^3}}}{\rho ^2}d\rho } \int\limits_0^\pi {\sin \theta d\theta } .\]
As it can be seen, the triple integral is transformed into the product of three single-valued integrals, each of which can be calculated independently. As a result, we obtain
\[
I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{e^{{\rho ^3}}}{\rho ^2}d\rho } \int\limits_0^\pi {\sin \theta d\theta }
= \left[ {\left. \varphi \right|_0^{2\pi }} \right] \cdot \int\limits_0^1 {\left( {{e^{{\rho ^3}}} \cdot \frac{1}{3}d{\rho ^3}} \right)} \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right]
= 2\pi \cdot \frac{1}{3}\left[ {\left. {\left( {{e^{{\rho ^3}}}} \right)} \right|_{{\rho ^3} = 0}^{{\rho ^3} = 1}} \right] \cdot \left( { - \cos \pi + \cos 0} \right)
= \frac{{2\pi }}{3} \cdot \left( {e - 1} \right) \cdot 2
= \frac{{4\pi }}{3}\left( {e - 1} \right).\]
See more problems on Page 2.
